Prove by induction that #f(n)=2^(2n-1)+3^(2n-1)# is divisible by 5 for #n in ZZ^+#?

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Answer 1 · 2018-01-28T16:44:29

See below.

Note that for #m# odd we have

#(a^m+b^m)/(a+b) = a^(m-1)-a^(m-2)b+ a^(m-3)b^2 + cdots -a b^(m-2)+b^(m-1)#

which demonstrates the afirmation.

Now by finite induction.

For #n = 1#

#2+3 = 5# which is divisible.

now supposing that

#2^(2n-1)+3^(2n-1)# is divisible we have

#2^(2(n+1)-1)+3^(2(n+1)-1) =2^(2n-1) 2^2+3^(2n-1)3^2=#
#= 2^(2n-1) 2^2+3^(2n-1)2^2+5 xx 3^(2n-1)=#

#= 2^2(2^(2n-1)+3^(2n-1))+5 xx 3^(2n-1)# which is divisible by #5#

so it is true.